Question: $f\,^{\prime}(x)=-5x^4+9x^2$ and $f(3)=-80$. $f(-3) = $
Finding $f(x)$ We have $f'(x)=-5x^4+9x^2$ and we want to find $f(x)$ : $\begin{aligned}f(x) &= \int f'(x)\,dx \\\\ & = \int (-5x^4+9x^2)\,dx \\\\ & = {-x^5+3x^3} {+ C} \end{aligned}$ Finding $ C$ Goal: We need to find $ C$ such that $f(3)=-80$. Here's what we get when we plug in $3$ : $\begin{aligned}f(3)&={-(3)^5+3(3)^3} {+ C}\\\\ &={-162} {+ C} \end{aligned}$ We are given that this must equal $-80$ : $-80 = {-162} {+ C}$ Solving the equation gives us ${C=82}$. Finding $f(-3)$ Now, we have that $f(x)={-x^5+3x^3} {+82}$. Let's find $f(-3)$ by plugging in $-3$ : $\begin{aligned}f(-3)&=-(-3)^5+3(-3)^3 + 82\\\\ &=244 \end{aligned}$ The answer $f(-3) = 244$